Show that if $T: \Bbb R^n \to \Bbb R^k$ is affine, then $T(x) = \lambda(x) + y_0$ for $\lambda : \Bbb R^n \to \Bbb R^k$ a linear map and $y_0 \in \Bbb R^k$ fixed.
I'm trying to understand the problem by simple examples and I considered $T: \Bbb R^2 \to \Bbb R^3$, then by definition $$T\left( \sum t_j p_j \right) = \sum t_jT(p_j)$$ where $\{p_0, p_1, \dots, p_m\} \subset \Bbb R^n$ is an affine independent set. In my case I have the standard basis $\{e_1, e_2\} \subset \Bbb R^2$ so $$T(t_1e_1+t_2e_2) = t_1T(e_1)+t_2T(e_2).$$ Now I should be able to show that $T(x) = \lambda(x) + y_0$, where $\lambda$ is a linear map $\Bbb R^2 \to \Bbb R^3$ and $y_0 \in \Bbb R^3$ a vector.
I don't quite get what this $\lambda$ should be. Somehow the intuition for this is that we should have a linear map which we then translate by $y_0$ in order to map the origin in the affine space to the corresponding origin in $\Bbb R^3$?
Based on the goal formula $T(x) = \lambda(x) + y_0$, we see that $T(0) = y_0$. So we really just have to show the map $\lambda(x) = T(x) - T(0)$ is linear. Let $x, y \in \mathbb{R}^n$ and $c \in \mathbb{R}$ be given.
For additivity of $\lambda$, use the fact that $1+1-1 = 1$ and the affineness of $T$ to get $$ \begin{align*} \lambda(x+y) & = T(x+y)-T(0) = T(x+y - 0)-T(0) = [T(x) + T(y) - T(0)]-T(0)\\ & = [T(x) - T(0)] + [T(y)-T(0)] = \lambda(x) + \lambda(y). \end{align*} $$ For scalar multiplicativity, use the fact that $c + (1-c) = 1$: $$ \begin{align*} \lambda(cx) & = T(cx) - T(0) = T(cx +(1-c)0) - T(0) = [cT(x) +(1-c)T(0)] - T(0)\\ & = cT(x) - cT(0) = c\lambda(x). \end{align*} $$ Thus $\lambda$ is linear, and by its definition, $T(x) = \lambda(x) + T(0)$ for all $x$, as desired.